Question

The eccentricity of the hyperbola $9x^2-{16y}^2+72x+32y-16=0$, is

• A. $\frac{5}{4}$
• B. $\frac{4}{5}$
• C. $\frac{9}{16}$
• D. $\frac{16}{9}$

Answer 1

The correct option is A $\frac{5}{4}$

We have $9(x^2+8x)-16(y^2-2y)=16$
$\Rightarrow 9{(x+4)}^2-16{(y-1)}^2=144$
$\Rightarrow \frac{{(x+4)}^2}{16}-\frac{{(y-1)}^2}{9}=1$
Here, $a^2=16,b^2=9$
$\therefore e^2=1+\frac{b^2}{a^2}=1+\frac{9}{16}=\frac{25}{16}$
$\Rightarrow e=\frac{5}{4}$
Hence, option 'A' is correct.